JEE Main 2025 Apr 7 Shift 1, Mathematics Q24: Algebra Of Matrices

When is the matrix singular? A $2\times 2$ matrix $\left(\begin{array}{c}ab\\ cd\end{array}\right)$ is singular when its determinant is zero:

$ad-bc=0\Rightarrow ad=bc$

So we need to count 4-tuples $(a,b,c,d)$ from $\{2,3,6,9\}$ where $ad$ and $bc$ give the same product.

Idea: Group the ordered pairs by their product. If a product $p$ can be made in $N$ ways, then $(a,d)$ has $N$ choices and $(b,c)$ has $N$ choices independently, giving $N^2$ singular matrices for that product. Add up over all products.

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Step 1: List all 16 products.

$\times$	$2$	$3$	$6$	$9$
$2$	$4$	$6$	$12$	$18$
$3$	$6$	$9$	$18$	$27$
$6$	$12$	$18$	$36$	$54$
$9$	$18$	$27$	$54$	$81$

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Step 2: Count ways to make each product.

Product $p$	Ways $N$	$N^2$
$4,9,36,81$	$1$ each	$1+1+1+1=4$
$6,12,27,54$	$2$ each	$4+4+4+4=16$
$18$	$4$ (since $2\times 9=3\times 6$)	$16$
Total	$16$ ✓	$36$

The highlighted product $18$ is the special case: four pairs make it, because $2\times 9$ and $3\times 6$ both equal $18$.

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Answer:

$\text{Singular matrices}=4+16+16=36$

$\therefore$ Answer = $\boxed{36}$.